What does an exact equation solution mean? I've only begun to study exact equations so this question will be naive.
I've went through the process of taking a given differential equation, showing that it's an exact equation and hence the derivative of some function $\psi$.  Figuring out the function $\psi$ is supposed to be a solution but I don't understand what is actually solved.  If I know $\psi$, what information do I have that I didn't have when I only had the differential equation?  What can I do with $\psi$?
 A: Let's say that you have the equation
$$
f(x,y)dx+g(x,y)dy=0
$$
and that it is exact. Then you find $\psi(x,y)$ such that
$$
\frac{\partial \psi}{\partial x}=f,\quad\frac{\partial \psi}{\partial y}=g.
$$
The solution is given in implicit form as
$$
\psi(x,y)=C.
$$
What does this mean? That from the implicit solution you can solve for $y$ in terms of $x$, and that the correspondig function $y=y(x,C)$ is a solution of the differential equation.
Example:
$$
2xdx+2ydy=0
$$
is exact. The solution is given in implicit form as
$$
x^2+y^2=C.
$$
From it you can get explicit solutions:
$$
y(x)=\sqrt{C-x^2},\quad C>0,\quad |x|<\sqrt C.
$$
This last step is not always possible (meaning that although a solution $y=y(x,C)$ exists, you may not be able to find an explicit solution in terms of elementary functions.)
A: Hint: Let you have an equation like $y'=f(x,y)$ which can be written as $y'=-\frac{M(x,y)}{N(x,y)}$ for some two functions $M(x,y)$ and $N(x,y)$. So you have $$\frac{dy}{dx}=-\frac{M(x,y)}{N(x,y)}$$ or $$M(x,y)dx+N(x,y)dy=0$$ If you can write the LHS of the equation as a differential of a certain $\phi$ such that $$d\phi=M(x,y)dx+N(x,y)dy$$, then by integration of both sides you get $$d\phi=0\Longrightarrow\phi=C$$ for some constant $C$.
